But in that case P(Awake|Heads) should, consistently with this interpretation, refer to your being awakened at all conditioned on the case where the coin landed heads. This is (1/3+2/3)/2 = 0.5 — Pierre-Normand

Me being awaked at all conditioned on the case where the coin lands heads is 1/3, given that if it lands heads then I am only woken up if I was assigned the number 1.

If you mean P(Awake) to refer to the probability of your being awakened at all (on at least one day) then P(Awake) is indeed 0.5. — Pierre-Normand

And it's the same in the normal case. The probability of being awakened at all (on at least one day) is 1. That's what should be used in Bayes' theorem.

Then this goes back to what I said above. These are two different questions with, I believe, two different answers:

1. If the experiment is run once, what is Sleeping Beauty's credence that the coin landed heads?

2. If the experiment is repeated several times, what is the probability that a randomly selected interview from the set of all interviews followed the coin landing heads? — Michael

I agree those are different btw. They describe completely different approaches to modelling the problem. That doesn't immediately tell us which SB ought to model the situation as, or whether they're internally coherent.

1. If the experiment is run once, what is Sleeping Beauty's credence that the coin landed heads?

2. If the experiment is repeated several times, what is the probability that a randomly selected interview from the set of all interviews followed the coin landing heads?

Thirders answer the second question, which I believe is the wrong answer to the first question. The experiment doesn't work by randomly selecting an interview from a set of interviews after repeating the experiment several times and then dropping Sleeping Beauty into it. — Michael

I would rather say that the experience works by ensuring that Sleeping Beauty finds herself being awoken in circumstances that she knows to be twice as likely to occur (because twice a frequent) as a result of a coin having landed heads than as a result of this coin having landed tails. This formulation entails no metaphysical queerness.

They describe completely different approaches to modelling the problem. That doesn't immediately tell us which SB ought to model the situation as, or whether they're internally coherent. — fdrake

One clue to this is to let SB bet on the outcome that her credence is about and see if her betting behavior leads her to realize the EV she is anticipating.

One clue to this is to let SB bet on the outcome that her credence is about and see if her betting behavior leads her to realize the EV she is anticipating. — Pierre-Normand

I never buy betting arguments unless the random variables are set up!

I never buy betting arguments unless the random variables are set up! — fdrake

They are!

Can you put them to me in excruciating detail please?

One clue to this is to let SB bet on the outcome that her credence is about and see if her betting behavior leads her to realize the EV she is anticipating. — Pierre-Normand

In my extreme example she wins in the long run (after 2¹⁰⁰ experiments) by betting on the coin landing heads 100 times in a row.

It doesn't then follow that when the experiment is run once that Sleeping Beauty's credence that the coin landed heads 100 times in a row should be greater than her credence that it didn't.

As I have said before, that she can bet more times if one outcome happens just isn't that that outcome is more probable.

I would rather say that the experience works by ensuring that Sleeping Beauty finds herself being awoken in circumstances that she knows to be twice as likely to occur (because twice a frequent) as a result of a coin having landed heads than as a result of this coin having landed tails. — Pierre-Normand

Then you have to say the same about my extreme example; that even when she knows that the experiment is only being run once, Sleeping Beauty's credence that the coin landed heads 100 times in a row should be greater than her credence that it didn't.

And I think that's an absurd conclusion, showing that your reasoning is faulty.

There is a space of possible awakening/interview events A that are being characterised by the day in which they occur ((M)onday or (T)uesday) and by the state of a coin that has been flipped prior to them occurring ((H)eads or (T)ails). P(H) = P(T) = 0.5. The rest of the conditional dependencies are part of the stipulation of the problem or can be inferred from them.

There is a space of possible awakening/interview events A that are being characterised by the day in which they occur ((M)onday or (T)uesday) and by the state of a coin that has been flipped prior to them occurring ((H)eads or (T)ails). P(H) = P(T) = 0.5. — Pierre-Normand

There's actually two spaces. See here.

Then you have to say the same about my extreme example. Even when she knows that the experiment is only being run once, Sleeping Beauty's credence that the coin landed heads 100 times in a row is greater than here credence that it didn't.

And I think that's an absurd conclusion, showing that your reasoning is false. — Michael

I'm not sure why you think this is absurd. Compare again my lottery study example. Suppose there are one billion people on the waiting list. If a coin lands heads 20 times in a row, then, 100 million people get pulled from the list. Else, one single person gets pulled from the list. I am then informed that I got pulled from the list (but not whether I am alone or one from 100 million). Is it absurd for me to believe that the coin landed heads 20 times in a row? My credence in this proposition should be roughly 99%

In Sleeping Beauty's case, your intuition that her credence in the high probability of the sequence of heads ought to be absurd apparently stems from your unwillingness to contemplate the possibility of her being able to updating it on the basis of her knowledge that she might be awoken an even more absurdly large number of times as a consequence of this very unlikely event. The need for her to update her credence upon being awakened stems from her lacks of power of distinguishing those events in respect of their causes (i.e. the coin flip results). Another reason why you are neglecting the need she has for updating her credence might be a result of your characterizing the experiment that is being run once as starting with a sequence of coin flips. But from Sleeping Beauty's perspective, it really begins with a random awakening.

I'm not sure why you think this is absurd. Compare again my lottery study example. Suppose there are one billion people on the waiting list. If a coin lands heads 20 times in a row, then, 100 million people get pulled from the list. Else, one single person gets pulled from the list. I am then informed that I got pulled from the list (but not whether I am alone or one from 100 million). Is it absurd for me to believe that the coin landed heads 20 times in a row? My credence in this proposition should be roughly 99% — Pierre-Normand

The difference is that the unconditional probability of being called up is very low, and so just being called up at all affects one's credence. In the Sleeping Beauty case (both the normal and my extreme version), she's guaranteed to be woken up either way.

In Sleeping Beauty's case, your intuition that her credence in the high probability of the sequence of heads ought to be absurd apparently stems from your unwillingness to contemplate the possibility of her being able to updating it on the basis of her knowledge that she might be awoken an even more absurdly large number of times as a consequence of this very unlikely event. — Pierre-Normand

She's certainly able to update it on the basis of her knowledge that she might be awoken an even more absurdly large number of times as a consequence of this very unlikely event. I'm saying that it's irrational of her to.

The only rational approach, upon waking, is to recognize that it landing heads 100 times in a row is so unlikely that it almost certainly didn't, and that this is her first and only interview.

There's actually two spaces. See here. — Michael

Yes, I agree with your representation.

The difference is that the unconditional probability of being called up is very low, and so just being called up at all affects one's credence. In the Sleeping Beauty case (both the normal and my extreme version), she's guaranteed to be awoken either way. — Michael

I can easily adjust my lottery study example such that I am guaranteed to be selected but, once selected, the very (unconditionally) low event that led to my selection (alongside a very large number of people) still is more likely than not to have been the cause of my selection. All that is needed is to shorten the waitlist by about 98%.

Regarding betting, expected values, and probability:

Rather than one person repeat the experiment 2¹⁰⁰ times, the experiment is done on 2¹⁰⁰ people, with each person betting that the coin will land heads 100 times in a row. 2¹⁰⁰ - 1 people lose, and 1 person wins, with the winner's winnings exceeding the sum of the losers' losses. The expected value is greater than the cost, but the probability of being a winner is still $1\over2^{100}$.

Even though I could win big, it is more rational to believe that I will lose.

It is a non sequitur to claim that a greater EV means a greater probability.

She's certainly able to update it on the basis of her knowledge that she might be awoken an even more absurdly large number of times as a consequence of this very unlikely event. I'm saying that it's irrational of her to.

The only rational approach, upon waking, is to recognize that it landing heads 100 times in a row is so unlikely that it almost certainly didn't, and that this is her first and only interview. — Michael

She does recognize that for the coin to land 100 times in a row is unconditionally unlikely. But why would it not be rational for her to condition her credence in the truth of this proposition on her knowledge that her awakening isn't an event that she can distinguish in respect of its cause, and that a very large number of such indistinguishable awakening events can stem from such an unlikely cause?

Regarding betting, expected values, and probability:

Rather than one person repeat the experiment 2^100 times, the experiment is done on 2^100 people, with each person betting that the coin will land heads 100 times in a row. 2^100 - 1 people lose, and 1 person wins, with the winner's winnings exceeding the sum of the losers' losses. The expected value of betting that the coin will land heads 100 is greater than the cost, but the probability of winning is still 1/2 ^100

Even though I could win big, it is more rational to believe that I will lose. — Michael

For sure, but your new variation doesn't mirror the Sleeping Beauty problem anymore. You earlier version was better. We must rather imagine that in the unlikely event that the coin lands heads 100 times in a row, 2^101 persons get pulled from the waitlist (or else, only one person does). In that case, if you are one random person who has been pulled from the waitlist, it is twice as likely that your have been so pulled as a result of the coin having landed heads 100 times in a row than not. It is hence rational for you to make this bet and in the long run 2/3 of the people pulled from the waitlist who make that bet with be right.

We must rather imagine that in the unlikely event that the coin lands heads 100 times in a row, 2^101 persons get pulled from the waitlist (or else, only one person does). In that case, if you are one random person who has been pulled from the waitlist, it is twice as likely that your have been so pulled as a result of the coin having landed heads 100 times in a row than not. It is hence rational for you to make this bet and in the long run 2/3 of the people pulled from the waitlist who make that bet with be right. — Pierre-Normand

This isn't comparable to the Sleeping Beaty problem because being a participant isn't guaranteed. That makes all the difference.

Compare these two scenarios:

1. There is a probability of 1 that I will be a participant if the coin lands heads 100 times and a probability of 1 that I will be a participant if the coin doesn't land heads 100 times

2. There is a probability of 1 that I will be a participant if the coin lands heads 100 times and a probability of $1\over2^{101}$ that I will be a participant if the coin doesn't land heads 100 times

In the first case it is rational to believe that the probability that the coin landed heads 100 times is $1\over2^{100}$

In the second case it is rational to believe that the probability that the coin landed heads 100 times is $2\over3$.

It certainly isn't the case that the two have the same answer. Surely you at least accept that? If so then the question is which of 1 and 2 properly represents the traditional problem. I say 1. There is a probability of 1 that she will be a participant if the coin lands heads and a probability of 1 that she will be a participant if the coin lands tails.

This isn't comparable to the Sleeping Beaty problem because being a participant isn't guaranteed. That makes all the difference. — Michael

I appreciate your viewpoint, but I can modify my analogies to meet the condition in your first scenario.

Scenario 3 (Lottery study)

Imagine that tickets numbered from one to 2^100 are randomly mixed. We have a waitlist of 2^101 participants. The experimenter selects tickets one by one from the mixed pile. When they draw the 'winning' ticket numbered 2^100, they select (2^100)+1 participants from the waitlist and assign them this ticket. Otherwise, they select just one participant and allocate the losing ticket drawn. This protocol guarantees that the waitlist is exhausted, and everyone gets a chance to participate.

As one of the 2^101 participants, your credence that you've been assigned the winning ticket numbered 2^100 should be equal to (or slightly greater than) the probability that you haven't, as half the participants (plus one) have been assigned this ticket.

Scenario 4 (Sleeping Beauty)

Let's again take tickets numbered from one to 2^100, randomly mixed. We have a waitlist of 2^100 participants. Each participant is assigned a ticket, put to sleep, and the participant with the 'winning' ticket is awoken and interviewed (2^100)+1 times, while everyone else is awoken and interviewed only once.

As one of the 2^100 participants who has just been awoken, your credence that you've been assigned the winning ticket numbered 2^100 should be equal to (or slightly greater than) the probability that you haven't, given that half the participant interviews (plus one) are with participants assigned the winning ticket. Or so I still would argue; here I am just addressing your most recent objection regarding guaranteed participation.

Scenario 5 (Bonus for @fdrake)

Returning to the original Sleeping Beauty problem: upon awakening and awaiting her interview, Sleeping Beauty reflects on her credence that the coin landed on heads. As she checks her smuggled-in cellphone and learns that today is Monday, her halfer position would require her to maintain her credence at 1/2, despite ruling out the possibility of it being Tuesday. The thirder position, on the other hand, allows her to adjust her credence from 2/3 to 1/2 after acquiring this new piece of information. This seems more plausible, and I would like the halfer to explain why ruling out the Tuesday scenario doesn't affect their credence in the coin toss outcome at all.

I would like the halfer to explain why ruling out the Tuesday scenario doesn't affect their credence in the coin toss outcome at all. — Pierre-Normand

For Lewis, if I recall correctly, it raises her credence for heads from 1/2 to 2/3, which he finds curious, but that's it.

I now realize that in the OP's stipulation of the problem, and in line with most discussions of it, it is when the fair coin lands tails that Sleeping Beauty is awoken twice, and else she is awoken once. I all my examples I had assumed the opposite. This is just an arbitrary convention but I hope my unfortunate reversal of it didn't generated any confusion.

In any case, in the future, I'll revert back to the conventional practice.

I would like the halfer to explain why ruling out the Tuesday scenario doesn't affect their credence in the coin toss outcome at all. — Pierre-Normand

Well, consider the Venn diagram here (which you said you agreed with).

There are two probability spaces. Monday or Tuesday is one consideration and Heads or Tails is a second consideration. Finding out that today is Monday just removes the blue circle.

As one of the 2^100 participants who has just been awoken, your credence that you've been assigned the winning ticket numbered 2^100 should be equal to (or slightly greater than) the probability that you haven't, given that half the participant interviews (plus one) are with participants assigned the winning ticket. — Pierre-Normand

I think this is a non sequitur. That most interviews are with a winner isn't that I am mostly likely a winner. Rather, if most participants are a winner then I am most likely a winner.

So there are two ways for the participants to approach the problem:

1. I should reason as if I am randomly selected from the set of all participants
2. I should reason as if my interview is randomly selected from the set of all interviews

I don't think the second is the rational approach. The experiment doesn't work by randomly selecting from the set of all interviews and then dropping each participant into that interview (in such a case the second might be the more rational approach, although how it would be determined which participant(s) are chosen to have more than one interview and the order in which the participants are placed is unclear, and it may be that there is more than one winner). The experiment works by giving each participant a ticket (and in such a case I think the first is the more rational approach).

Finding out that today is Monday just removes the blue circle. — Michael

I agree with the idea that Sleeping Beauty's credence in H is updated to 1/2 after she learns that her current awakening is occurring on a Monday. The question is what was this credence updated from, before she learned that the current day was Monday. At that point she could not rule out that the current day was Tuesday. Does that not imply that her credence in H was lower than 1/2 before she learned that the day was Monday? Before she learned that, all three mutually exclulsive areas of the Venn diagram represent possible states for her to be in. She therefore should have non-zero credences for each one of them.

I agree with the idea that Sleeping Beauty's credence in H is updated to 1/2 after she learns that her current awakening is occurring on a Monday. The question is what was this credence updated from, before she learned that the current day was Monday. At that point she could not rule out that the current day was Tuesday. Does that not imply that her credence in H was lower than 1/2 before she learned that the day was Monday? Before she learned that, all three mutually exclulsive areas of the Venn diagram represent possible states for her to be in. She therefore should have non-zero credences for each one of them. — Pierre-Normand

There are two approaches. The normal halfer approach is:

P(Heads) = 1/2
P(Tails) = 1/2
P(Monday) = 3/4
P(Tuesday) = 1/4

The double halfer approach is:

P(Heads) = 1/2
P(Tails) = 1/2
P(Monday) = 1/2
P(Tuesday) = 1/2

The double halfer approach does entail:

P(Heads & Monday) = P(Tails & Monday) = P(Tails & Tuesday) = 1/2

This reflects what I said before:

The probability that the coin will land heads and she will be woken on Monday is 1/2.
The probability that the coin will land tails and she will be woken on Monday is 1/2.
The probability that the coin will land tails and she will be woken on Tuesday is 1/2.

As the Venn diagram shows, there are two (overlapping) probability spaces, hence why the sum of each outcome's probability is greater than 1.

I'm undecided on whether or not I'm a double halfer.

The double halfer approach does entail:

P(Heads & Monday) = P(Tails & Monday) = P(Tails and Tuesday) = 1/2 — Michael

From the external point of view of the experimenter, it makes sense that the tree probabilities add up to more than one since the three outcomes are non exclusive. The events (Tails & Monday) and (Tails and Tuesday) can (and indeed must) be realized jointly. The intersection zone in the Venn diagram highlights the fact that those two events aren't mutually exclusive. The perspective (and epistemic position) of Sleeping Beauty is different. When she is awoken, those three possible states are mutually exclusive since it is not possible for her to believe that, in her current state of awakening, the current day is both Monday and Tuesday. Her credences in the three mutually exclusive states that she can be in therefore must add up to 1.

From the external point of view of the experimenter, it makes sense that the tree probabilities add up to more than one since the three outcomes are non exclusive. The events (Tails & Monday) and (Tails and Tuesday) can (and indeed must) be realized jointly. The intersection zone in the Venn diagram highlights the fact that those two events aren't mutually exclusive. The perspective (and epistemic position) of Sleeping Beauty is different. When she is awoken, those three possible states are mutually exclusive since it is not possible for her to believe that, in her current state of awakening, the current day is both Monday and Tuesday. Her credences in the three mutually exclusive states that she can be in therefore must add up to 1. — Pierre-Normand

I think this is best addressed by a variation I described here:

There are four people, each assigned a number (unknown to them) between 1 and 4. Two of them are to be put to sleep at random, determined by a single coin toss: if heads then 1 and 2 are put to sleep; if tails then 3 and 4 are put to sleep.

After the coin toss one of the two is put to sleep first. For each of the remaining three, what is the probability that they will be put to sleep? Is it 1/3, because there are three of them and only one is to be put to sleep, or is it 1/2, because that was the probability before the first was put to sleep?

Before anyone was put to sleep, for each of the four participants the probability of being put to sleep was 1/2, which of course doesn't add to 1.

And depending on your answer to this scenario, for each of the remaining three participants the probability of being put to sleep is 1/2.

So there are two ways for the participants to approach the problem:

1. I should reason as if I am randomly selected from the set of all participants
2. I should reason as if my interview is randomly selected from the set of all interviews — Michael

To apply this to the traditional problem: there are two Sleeping Beauties; one will be woken on Monday and one on both Monday and Tuesday, determined by a coin toss.

What is their credence that they have been or will be woken twice?

Do they reason as if they are randomly selected from the set of all participants or do they reason as if their interview is randomly selected from the set of all interviews?

Which is the most rational?

The math can prove that the former reasoning gives an answer of $1\over2$ and the latter an answer of $1\over3$ but I don’t think it can prove which reasoning is the correct to use. It might be that there is no correct answer, only a more compelling answer.

Given the way the experiment is conducted I find the former reasoning more compelling. This is especially so with the example of the coin landing heads 100 times in a row.

I think this is basically a Monty Hall problem. I would say that the probability that I will be put to sleep is 1/2, that the probability that the person to my left will be put to sleep is 1/2, that the probability that the person to my right will be put to sleep is 1/2, and that the probability that one of the other two will be put to sleep is 1/2. — Michael

Your variation of the problem indeed appears to me to contain elements reminiscent of the Monty Hall problem, but with a key difference in the selection process.

In the original Monty Hall scenario, Monty's action of revealing a goat behind a door you didn't choose is non-random. This non-random revelation influences your strategy, prompting you to switch doors. (The probabilities of your initial choice and the remaining door hiding the car that were 1/3 and 1/3, respectively, get updated to 1/3 and 2/3.

Your scenario can be compared to a modified Monty Hall scenario. In this modified scenario, imagine Monty picks one of the unchosen doors at random and reveals a goat by chance. Because this revelation is a random event, it doesn't provide the same targeted information as in the traditional Monty Hall scenario. Consequently, the initial probability estimates (1/3) for your chosen door and the remaining unchosen door hiding a car are updated to 1/2 each. Therefore, you would be indifferent about whether to stick with your original door or switch to the remaining door.

Similarly, in your variation, if a participant is revealed in random way to have been selected to be put to sleep, much like Monty randomly revealing a goat, the probabilities for the remaining three participants (including yourself) would need to be updated similarly. If it's known that exactly one of the remaining three participants had been selected, then the probabilities for each of those three would increase to 1/3.

In this modified scenario, imagine Monty picks one of the unchosen doors at random and reveals a goat by chance. — Pierre-Normand

It's not exactly comparable as in my example he can only put to sleep one of the two people who will be put to sleep; he cannot put to sleep someone who won't be put to sleep.

A more comparable example would be if there are four doors, two containing a goat and two containing a car. You pick a door (say 1) and then Monty opens one of the two doors that contain a goat (say 2). What is the probability that your chosen door contains a car? What is the probability that the car is behind door 3, or door 4?